Solutions and Colligative Properties 2 Question 1
1. A solution is prepared by dissolving $0.6 \mathrm{~g}$ of urea (molar mass $=60 \mathrm{~g} \mathrm{~mol}^{-1}$ ) and $1.8 \mathrm{~g}$ of glucose (molar mass $=180 \mathrm{~g}$ $\left.\mathrm{mol}^{-1}\right)$ in $100 \mathrm{~mL}$ of water at $27^{\circ} \mathrm{C}$. The osmotic pressure of the solution is $\left(R=0.08206 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$
(2019 Main, 12 April II)
(a) $8.2 \mathrm{~atm}$
(b) $2.46 \mathrm{~atm}$
(c) $4.92 \mathrm{~atm}$
(d) $1.64 \mathrm{~atm}$
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Solution:
- Key Idea Osmotic pressure is proportional to the molarity $(C)$ of the solution at a given temperature (T).
Thus, $\pi \propto C, \pi=C R T$ (for dilute solution)
$$ \pi=\frac{n}{V} R T $$
For the relation, $\pi=C R T=\frac{n}{V} R T$
Given, mass of urea $=0.6 \mathrm{~g}$
Molar mass of urea $=60 \mathrm{~g} \mathrm{~mol}^{-1}$
Mass of glucose $=1.8 \mathrm{~g}$
Molar mass of glucose $=180 \mathrm{~g} \mathrm{~mol}^{-1}$
$$ \begin{aligned} \pi & =\frac{\left.\left[n_{2} \text { (urea }\right)+n_{2}(\text { glucose })\right]}{V} R T \ & =\frac{\left(\frac{0.6}{60}+\frac{1.8}{180}\right)}{100} \times 1000 \times 0.0821 \times 300 \ & =(0.01+0.01) \times 10 \times 0.0821 \times 300 \ \pi & =4.92 \mathrm{~atm} \end{aligned} $$
